mac2312 ch 3 solutions

November 10, 2008 21:04 ”SSM ET chapter 3” Sheet number 1 Page number 54 black

54

CHAPTER 3

Topics in Differentiation

EXERCISE SET 3.1

1. (a) 1 + y + xdy

dx− 6x2 = 0,

dy

dx=

6x2 − y − 1x

(b) y =2 + 2x3 − x

x=

2x

+ 2x2 − 1,dy

dx= − 2

x2+ 4x

(c) From Part (a),dy

dx= 6x− 1

x− 1xy = 6x− 1

x− 1x

(2x

+ 2x2 − 1)

= 4x− 2x2

3. 2x+ 2ydy

dx= 0 so

dy

dx= −x

y

5. x2 dy

dx+ 2xy + 3x(3y2)

dy

dx+ 3y3 − 1 = 0

(x2 + 9xy2)dy

dx= 1− 2xy − 3y3 so

dy

dx=

1− 2xy − 3y3

x2 + 9xy2

7. − 12x3/2

−dydx

2y3/2= 0,

dy

dx= −y

3/2

x3/2

9. cos(x2y2)[x2(2y)

dy

dx+ 2xy2

]= 1,

dy

dx=

1− 2xy2 cos(x2y2)2x2y cos(x2y2)

11. 3 tan2(xy2 + y) sec2(xy2 + y)(

2xydy

dx+ y2 +

dy

dx

)= 1

sody

dx=

1− 3y2 tan2(xy2 + y) sec2(xy2 + y)3(2xy + 1) tan2(xy2 + y) sec2(xy2 + y)

13. 4x− 6ydy

dx= 0,

dy

dx=

2x3y, 4− 6

(dy

dx

)2

− 6yd2y

dx2= 0,

d2y

dx2= −

3(dydx

)2

− 2

3y=

2(3y2 − 2x2)9y3

= − 89y3

15.dy

dx= −y

x,d2y

dx2= −x(dy/dx)− y(1)

x2= −x(−y/x)− y

x2=

2yx2

17.dy

dx= (1 + cos y)−1,

d2y

dx2= −(1 + cos y)−2(− sin y)

dy

dx=

sin y(1 + cos y)3

19. By implicit differentiation, 2x + 2y(dy/dx) = 0,dy

dx= −x

y; at (1/2,

√3/2),

dy

dx= −

√3/3; at

(1/2,−√

3/2),dy

dx= +

√3/3. Directly, at the upper point y =

√1− x2,

dy

dx=

−x√1− x2

=

− 1/2√3/4

= −1/√

3 and at the lower point y = −√

1− x2,dy

dx=

x√1− x2

= +1/√

3.

21. false; x = y2 defines two functions y = ±√x. See Definition 3.1.1.

23. false; the equation is equivalent to x2 = y2 which is satisfied by y = |x|.


Exercise Set 3.1 55

25. 4x3 + 4y3 dy

dx= 0, so

dy

dx= −x

3

y3= − 1

153/4≈ −0.1312.

27. 4(x2 + y2)(

2x+ 2ydy

dx

)= 25

(2x− 2y

dy

dx

),

dy

dx=x[25− 4(x2 + y2)]y[25 + 4(x2 + y2)]

; at (3, 1)dy

dx= −9/13

29. 2x + xdy

dx+ y + 2y

dy

dx= 0. Substitute y = −2x to obtain −3x

dy

dx= 0. Since x = ±1 at the

indicated points,dy

dx= 0 there.

31. (a)

–4 4

–2

2

x

y

(b) Implicit differentiation of the curve yields (4y3 + 2y)dy

dx= 2x− 1 so

dy

dx= 0

only if x = 1/2 but y4 + y2 ≥ 0 so x = 1/2 is impossible..

(c) x2 − x− (y4 + y2) = 0, so by the Quadratic Formula, x =−1±

√(2y2 + 1)2

2= 1 + y2,−y2,

and we have the two parabolas x = −y2, x = 1 + y2.

33. Solve the simultaneous equations y = x, x2−xy+y2 = 4 to get x2−x2+x2 = 4, x = ±2, y = x = ±2,so the points of intersection are (2, 2) and (−2,−2).

By implicit differentiation,dy

dx=y − 2x2y − x

. When x = y = 2,dy

dx= −1; when x = y = −2,

dy

dx= −1;

the slopes are equal.

35. The point (1,1) is on the graph, so 1 + a = b. The slope of the tangent line at (1,1) is −4/3; use

implicit differentiation to getdy

dx= − 2xy

x2 + 2ayso at (1,1), − 2

1 + 2a= −4

3, 1 + 2a = 3/2, a = 1/4

and hence b = 1 + 1/4 = 5/4.

37. We shall find when the curves intersect and check that the slopes are negative reciprocals. For theintersection solve the simultaneous equations x2 + (y − c)2 = c2 and (x− k)2 + y2 = k2 to obtain

cy = kx =12

(x2 + y2). Thus x2 + y2 = cy + kx, or y2 − cy = −x2 + kx, andy − cx

= −x− ky

.

Differentiating the two families yields (black)dy

dx= − x

y − c, and (gray)

dy

dx= −x− k

y. But it was

proven that these quantities are negative reciprocals of each other.


56 Chapter 3

39. (a)

–3 –1 2

–3

–1

2

x

y

(b) x ≈ 0.84(c) Use implicit differentiation to get dy/dx = (2y−3x2)/(3y2−2x), so dy/dx = 0 if y = (3/2)x2.

Substitute this into x3 − 2xy + y3 = 0 to obtain 27x6 − 16x3 = 0, x3 = 16/27, x = 24/3/3and hence y = 25/3/3.

41. Implicit differentiation of the equation of the curve yields rxr−1 +ryr−1 dy

dx= 0. At the point (1, 1)

this becomes r + rdy

dx= 0,

dy

dx= −1.

EXERCISE SET 3.2

1.1

5x(5) =

1x

3.1

1 + x5.

1x2 − 1

(2x) =2x

x2 − 1

7.d

dxlnx− d

dxln(1 + x2) =

1x− 2x

1 + x2=

1− x2

x(1 + x2)

9.d

dx(2 lnx) = 2

d

dxlnx =

2x

11.12

(lnx)−1/2

(1x

)=

12x√

lnx

13. lnx+ x1x

= 1 + lnx 15. 2x log2(3− 2x) +−2x2

(ln 2)(3− 2x)

17.2x(1 + log x)− x/(ln 10)

(1 + log x)219.

1lnx

(1x

)=

1x lnx

21.1

tanx(sec2 x) = secx cscx 23. − 1

xsin(lnx)

25.1

ln 10 sin2 x(2 sinx cosx) = 2

cotxln 10

27.d

dx

[3 ln(x− 1) + 4 ln(x2 + 1)

]=

3x− 1

+8x

x2 + 1=

11x2 − 8x+ 3(x− 1)(x2 + 1)

29.d

dx

[ln cosx− 1

2ln f(4− 3x2)

]= − tanx+

3x4− 3x2

31. true

33. if x > 0 thend

dxln |x| = 1/x; if x < 0 then

d

dxln |x| = 1/x, true


Exercise Set 3.2 57

35. ln |y| = ln |x|+ 13

ln |1 + x2|, dydx

= x3√

1 + x2

[1x

+2x

3(1 + x2)

]

37. ln |y| = 13

ln |x2 − 8|+ 12

ln |x3 + 1| − ln |x6 − 7x+ 5|

dy

dx=

(x2 − 8)1/3√x3 + 1

x6 − 7x+ 5

[2x

3(x2 − 8)+

3x2

2(x3 + 1)− 6x5 − 7x6 − 7x+ 5

]

39. (a) logx e =ln elnx

=1

lnx,d

dx[logx e] = − 1

x(lnx)2

(b) logx 2 =ln 2lnx

,d

dx[logx 2] = − ln 2

x(lnx)2

40. (a) From loga b =ln bln a

for a, b > 0 it follows that log(1/x) e =ln e

ln(1/x)= − 1

lnx, hence

d

dx

[log(1/x) e

]=

1x(lnx)2

(b) log(ln x) e =ln e

ln(lnx)=

1ln(lnx)

, sod

dxlog(ln x) e = − 1

(ln(lnx))2

1x lnx

= − 1x(lnx)(ln(lnx))2

41. f ′(x0) =1x0

= e, y − e−1 = e(x− x0) = ex− 1, y = ex− 1 +1e

43. f(x0) = f(−e) = 1, f ′(x)∣∣∣∣x=−e

= − 1e ,

y − 1 = − 1e (x+ e), y = − 1

ex

45. (a) Let the equation of the tangent line be y = mx and suppose that it meets the curve at

(x0, y0). Then m =1x

∣∣∣∣x=x0

=1x0

and y0 = mx0 + b = lnx0. So m =1x0

=lnx0

x0and

lnx0 = 1, x0 = e,m = 1e and the equation of the tangent line is y =

1ex.

(b) Let y = mx+ b be a line tangent to the curve at (x0, y0). Then b is the y-intercept and the

slope of the tangent line is m =1x0

. Moreover, at the point of tangency, mx0 + b = lnx0 or

1x0x0 + b = lnx0, b = lnx0 − 1, as required.

47. The area of the triangle PQR, given by |PQ||QR|/2 is required.|PQ| = w, and, by Exercise 45 Part (b), |QR| = 1, so area = w/2.

2

–2

1

x

y

P (w, ln w)Q

R

w

49. If x = 0 then y = ln e = 1, anddy

dx=

1x+ e

. But ey = x+ e, sody

dx=

1ey

= e−y. .


58 Chapter 3

51. Let y = ln(x+a). Following Exercise 49 we getdy

dx=

1x+ a

= e−y, and when x = 0, y = ln(a) = 0

if a = 1, so let a = 1, then y = ln(x+ 1).

53. (a) Set f(x) = ln(1 + 3x). Then f ′(x) =3

1 + 3x, f ′(0) = 3. But

f ′(0) = limx→0

f(x)− f(0)x

= limx→0

ln(1 + 3x)x

(b) f(x) = ln(1−5x), f ′(x) =−5

1− 5x, f ′(0) = −5. But f ′(0) = lim

x→0

f(x)− f(0)x

= limx→0

ln(1− 5x)x

55. (a) Let f(x) = ln(cosx), then f(0) = ln(cos 0) = ln 1 = 0, so f ′(0) = limx→0

f(x)− f(0)x

=

limx→0

ln(cosx)x

, and f ′(0) = − tan 0 = 0.

(b) Let f(x) = x√

2, then f(1) = 1, so f ′(1) = limh→0

f(1 + h)− f(1)h

= limh→0

(1 + h)√

2 − 1h

, and

f ′(x) =√

2x√

2−1, f ′(1) =√

2.

EXERCISE SET 3.3

1. (a) f ′(x) = 5x4 + 3x2 + 1 ≥ 1 so f is one-to-one on −∞ < x < +∞.

(b) f(1) = 3 so 1 = f−1(3);d

dxf−1(x) =

1f ′(f−1(x))

, (f−1)′(3) =1

f ′(1)=

19

3. f−1(x) =2x− 3, so directly

d

dxf−1(x) = − 2

x2. Using Formula (2),

f ′(x) =−2

(x+ 3)2, so

1f ′(f−1(x))

= −(1/2)(f−1(x) + 3)2,

d

dxf−1(x) = −(1/2)

(2x

)2

= − 2x2

5. (a) f ′(x) = 2x + 8; f ′ < 0 on (−∞,−4) and f ′ > 0 on (−4,+∞); not enough information. Byinspection, f(1) = 10 = f(−9), so not one-to-one

(b) f ′(x) = 10x4 + 3x2 + 3 ≥ 3 > 0; f ′(x) is positive for all x, so f is one-to-one(c) f ′(x) = 2 + cosx ≥ 1 > 0 for all x, so f is one-to-one(d) f ′(x) = −(ln 2)

(12

)x< 0 because ln 2 > 0, so f is one-to-one for all x.

7. y = f−1(x), x = f(y) = 5y3 + y − 7,dx

dy= 15y2 + 1,

dy

dx=

115y2 + 1

;

check: 1 = 15y2 dy

dx+dy

dx,dy

dx=

115y2 + 1

9. y = f−1(x), x = f(y) = 2y5 + y3 + 1,dx

dy= 10y4 + 3y2,

dy

dx=

110y4 + 3y2

;

check: 1 = 10y4 dy

dx+ 3y2 dy

dx,dy

dx=

110y4 + 3y2

11. Let P (a, b) be given, not on the line y = x. Let Q be its reflection across the line y = x, yet to bedetermined. Let Q1 have coordinates (b, a).


Exercise Set 3.3 59

(a) Since P does not lie on y = x, we have a 6= b, i.e. P 6= Q1 since they have different abscissas.The line ~PQ1 has slope (b− a)/(a− b) = −1 which is the negative reciprocal of m = 1 andso the two lines are perpendicular.

(b) Let (c, d) be the midpoint of the segment PQ1. Then c = (a + b)/2 and d = (b + a)/2 soc = d and the midpoint is on y = x.

(c) Let Q(c, d) be the reflection of P through y = x. By definition this means P and Q lie on aline perpendicular to the line y = x and the midpoint of P and Q lies on y = x.

(d) Since the line through P and Q1 is perpendicular to the line y = x it is parallel to the linethrough P and Q; since both pass through P they are the same line. Finally, since themidpoints of P and Q1 and of P and Q both lie on y = x, they are the same point, andconsequently Q = Q1.

13. If x < y then f(x) ≤ f(y) and g(x) ≤ g(y); thus f(x)+g(x) ≤ f(y)+g(y). Moreover, g(x) ≤ g(y),so f(g(x)) ≤ f(g(y)). Note that f(x)g(x) need not be increasing, e.g. f(x) = g(x) = x, bothincreasing for all x, yet f(x)g(x) = x2, not an increasing function.

15. 7e7x 17. x3ex + 3x2ex = x2ex(x+ 3)

19.dy

dx=

(ex + e−x)(ex + e−x)− (ex − e−x)(ex − e−x)(ex + e−x)2

=(e2x + 2 + e−2x)− (e2x − 2 + e−2x)

(ex + e−x)2= 4/(ex + e−x)2

21. (x sec2 x+ tanx)ex tan x 23. (1− 3e3x)e(x−e3x)

25.(x− 1)e−x

1− xe−x=

x− 1ex − x

27. f ′(x) = 2x ln 2; y = 2x, ln y = x ln 2,1yy′ = ln 2, y′ = y ln 2 = 2x ln 2

29. f ′(x) = πsin x(lnπ) cosx;

y = πsin x, ln y = (sinx) lnπ,1yy′ = (lnπ) cosx, y′ = πsin x(lnπ) cosx

31. ln y = (lnx) ln(x3 − 2x),1y

dy

dx=

3x2 − 2x3 − 2x

lnx+1x

ln(x3 − 2x),

dy

dx= (x3 − 2x)ln x

[3x2 − 2x3 − 2x

lnx+1x

ln(x3 − 2x)]

33. ln y = (tanx) ln(lnx),1y

dy

dx=

1x lnx

tanx+ (sec2 x) ln(lnx),

dy

dx= (lnx)tan x

[tanxx lnx

+ (sec2 x) ln(lnx)]


60 Chapter 3

35. ln y = (lnx)(ln(lnx)),dy/dx

y= (1/x)(ln(lnx)) + (lnx)

1/xlnx

= (1/x)(1 + ln(lnx))

dy/dx =1x

(lnx)ln x(1 + ln lnx)

37.3√

1− (3x)2=

3√1− 9x2

39.1√

1− 1/x2(−1/x2) = − 1

|x|√x2 − 1

41.3x2

1 + (x3)2=

3x2

1 + x643. y = 1/ tanx = cotx, dy/dx = − csc2 x

45.ex

|x|√x2 − 1

+ ex sec−1 x 47. 0

49. 0 51. − 11 + x

(12x−1/2

)= − 1

2(1 + x)√x

53. false; y = Aex also satisfiesdy

dx= y

55. true; examine the cases x > 0 and x < 0 separately

57. (a) Let x = f(y) = cot y, 0 < y < π,−∞ < x < +∞. Then f is differentiable and one-to-oneand f ′(f−1(x)) = − csc2(cot−1 x) = −x2 − 1 6= 0, andd

dx[cot−1 x]

∣∣∣∣x=0

= limx→0

1f ′(f−1(x))

= − limx→0

1x2 + 1

= −1.

(b) If x 6= 0 then, from Exercise 48(a) of Section 0.4,d

dxcot−1 x =

d

dxtan−1 1

x= − 1

x2

11 + (1/x)2

= − 1x2 + 1

. For x = 0, Part (a) shows the same;

thus for −∞ < x < +∞, ddx

[cot−1 x] = − 1x2 + 1

.

(c) For −∞ < u < +∞, by the chain rule it follows thatd

dx[cot−1 u] = − 1

u2 + 1du

dx.

59. x3 + x tan−1 y = ey, 3x2 +x

1 + y2y′ + tan−1 y = eyy′, y′ =

(3x2 + tan−1 y)(1 + y2)(1 + y2)ey − x

61. (a) f(x) = x3 − 3x2 + 2x = x(x− 1)(x− 2) so f(0) = f(1) = f(2) = 0 thus f is not one-to-one.

(b) f ′(x) = 3x2 − 6x + 2, f ′(x) = 0 when x =6±√

36− 246

= 1 ±√

3/3. f ′(x) > 0 (f is

increasing) if x < 1−√

3/3, f ′(x) < 0 (f is decreasing) if 1−√

3/3 < x < 1 +√

3/3, so f(x)takes on values less than f(1−

√3/3) on both sides of 1−

√3/3 thus 1−

√3/3 is the largest

value of k.

63. (a) f ′(x) = 4x3 + 3x2 = (4x+ 3)x2 = 0 only at x = 0. But on [0, 2], f ′ has no sign change, so fis one-to-one.

(b) F ′(x) = 2f ′(2g(x))g′(x) so F ′(3) = 2f ′(2g(3))g′(3). By inspection f(1) = 3, sog(3) = f−1(3) = 1 and g′(3) = (f−1)′(3) = 1/f ′(f−1(3)) = 1/f ′(1) = 1/7 becausef ′(x) = 4x3 + 3x2. Thus F ′(3) = 2f ′(2)(1/7) = 2(44)(1/7) = 88/7.F (3) = f(2g(3)) = f(2·1) = f(2) = 25, so the line tangent to F (x) at (3, 25) has the equationy − 25 = (88/7)(x− 3), y = (88/7)x− 89/7.


Exercise Set 3.4 61

65. y = Aekt, dy/dt = kAekt = k(Aekt) = ky

67. (a) y′ = −xe−x + e−x = e−x(1− x), xy′ = xe−x(1− x) = y(1− x)

(b) y′ = −x2e−x2/2 + e−x

2/2 = e−x2/2(1− x2), xy′ = xe−x

2/2(1− x2) = y(1− x2)

69. ln y = ln 60− ln(5 + 7e−t),y′

y=

7e−t

5 + 7e−t=

7e−t + 5− 55 + 7e−t

= 1− 112y, so

dy

dt= r

(1− y

K

)y, with r = 1,K = 12.

71. f(x) = e3x, f ′(0) = limx→0

f(x)− f(0)x− 0

= 3e3x]x=0

= 3

73. limh→0

10h − 1h

=d

dx10x∣∣∣∣x=0

=d

dxex ln 10

∣∣∣∣x=0

= ln 10

75. lim∆x→0

9[sin−1(√

32 + ∆x)]2 − π2

∆x=

d

dx(3 sin−1 x)2

∣∣∣∣x=√

3/2

= 2(3 sin−1 x)3√

1− x2

∣∣∣∣x=√

3/2

= 2(3π

3)

3√1− (3/4)

= 12π

EXERCISE SET 3.4

1.dy

dt= 3

dx

dt

(a)dy

dt= 3(2) = 6 (b) −1 = 3

dx

dt,dx

dt= −1

3

3. 8xdx

dt+ 18y

dy

dt= 0

(a) 81

2√

2· 3 + 18

13√

2dy

dt= 0,

dy

dt= −2 (b) 8

(13

)dx

dt− 18

√5

9· 8 = 0,

dx

dt= 6√

5

5. (b) A = x2 (c)dA

dt= 2x

dx

dt

(d) FinddA

dt

∣∣∣∣x=3

given thatdx

dt

∣∣∣∣x=3

= 2. From Part (c),dA

dt

∣∣∣∣x=3

= 2(3)(2) = 12 ft2/min.

7. (a) V = πr2h, sodV

dt= π

(r2 dh

dt+ 2rh

dr

dt

).

(b) FinddV

dt

∣∣∣∣h=6,r=10

given thatdh

dt

∣∣∣∣h=6,r=10

= 1 anddr

dt

∣∣∣∣h=6,r=10

= −1. From Part (a),

dV

dt

∣∣∣∣h=6,r=10

= π[102(1) + 2(10)(6)(−1)] = −20π in3/s; the volume is decreasing.

9. (a) tan θ =y

x, so sec2 θ

dθ

dt=xdy

dt− y dx

dtx2

,dθ

dt=

cos2 θ

x2

(xdy

dt− y dx

dt

)


62 Chapter 3

(b) Finddθ

dt

∣∣∣∣x=2,y=2

given thatdx

dt

∣∣∣∣x=2,y=2

= 1 anddy

dt

∣∣∣∣x=2,y=2

= −14

.

When x = 2 and y = 2, tan θ = 2/2 = 1 so θ =π

4and cos θ = cos

π

4=

1√2

. Thus

from Part (a),dθ

dt

∣∣∣∣x=2,y=2

=(1/√

2)2

22

[2(−1

4

)− 2(1)

]= − 5

16rad/s; θ is decreasing.

11. Let A be the area swept out, and θ the angle through which the minute hand has rotated.

FinddA

dtgiven that

dθ

dt=

π

30rad/min; A =

12r2θ = 8θ, so

dA

dt= 8

dθ

dt=

4π15

in2/min.

13. Finddr

dt

∣∣∣∣A=9

given thatdA

dt= 6. From A = πr2 we get

dA

dt= 2πr

dr

dtso

dr

dt=

12πr

dA

dt. If A = 9

then πr2 = 9, r = 3/√π so

dr

dt

∣∣∣∣A=9

=1

2π(3/√π)

(6) = 1/√π mi/h.

15. FinddV

dt

∣∣∣∣r=9

given thatdr

dt= −15. From V =

43πr3 we get

dV

dt= 4πr2 dr

dtso

dV

dt

∣∣∣∣r=9

= 4π(9)2(−15) = −4860π. Air must be removed at the rate of 4860π cm3/min.

17. Finddx

dt

∣∣∣∣y=5

given thatdy

dt= −2. From x2 + y2 = 132 we get

2xdx

dt+ 2y

dy

dt= 0 so

dx

dt= −y

x

dy

dt. Use x2 + y2 = 169 to find that

x = 12 when y = 5 sodx

dt

∣∣∣∣y=5

= − 512

(−2) =56

ft/s.

13

x

y

19. Let x denote the distance from first base and y the distance

from home plate. Then x2 +602 = y2 and 2xdx

dt= 2y

dy

dt. When

x = 50 then y = 10√

61 sody

dt=x

y

dx

dt=

5010√

61(25) =

125√61

ft/s.

60 ft

y

x

Home

First

21. Finddy

dt

∣∣∣∣x=4000

given thatdx

dt

∣∣∣∣x=4000

= 880. From

y2 = x2 + 30002 we get 2ydy

dt= 2x

dx

dtso

dy

dt=x

y

dx

dt.

If x = 4000, then y = 5000 sody

dt

∣∣∣∣x=4000

=40005000

(880) = 704 ft/s.

Rocket

Camera

3000 ft

yx


Exercise Set 3.4 63

23. (a) If x denotes the altitude, then r − x = 3960, the radius of the Earth. θ = 0 at perigee, sor = 4995/1.12 ≈ 4460; the altitude is x = 4460 − 3960 = 500 miles. θ = π at apogee, sor = 4995/0.88 ≈ 5676; the altitude is x = 5676− 3960 = 1716 miles.

(b) If θ = 120◦, then r = 4995/0.94 ≈ 5314; the altitude is 5314− 3960 = 1354 miles. The rateof change of the altitude is given by

dx

dt=dr

dt=dr

dθ

dθ

dt=

4995(0.12 sin θ)(1 + 0.12 cos θ)2

dθ

dt.

Use θ = 120◦ and dθ/dt = 2.7◦/min = (2.7)(π/180) rad/min to get dr/dt ≈ 27.7 mi/min.

25. Finddh

dt

∣∣∣∣h=16

given thatdV

dt= 20. The volume of water in the

tank at a depth h is V =13πr2h. Use similar triangles (see figure)

to getr

h=

1024

so r =512h thus V =

13π

(512h

)2h =

25432

πh3,

dV

dt=

25144

πh2 dh

dt;dh

dt=

14425πh2

dV

dt,

dh

dt

∣∣∣∣h=16

=144

25π(16)2(20) =

920π

ft/min.

h

r

24

10

27. FinddV

dt

∣∣∣∣h=10

given thatdh

dt= 5. V =

13πr2h, but

r =12h so V =

13π

(h

2

)2h =

112πh3,

dV

dt=

14πh2 dh

dt,

dV

dt

∣∣∣∣h=10

=14π(10)2(5) = 125π ft3/min.

h

r

29. With s and h as shown in the figure, we want

to finddh

dtgiven that

ds

dt= 500. From the figure,

h = s sin 30◦ =12s so

dh

dt=

12ds

dt=

12

(500) = 250 mi/h.

sh

Ground

30°

31. Finddy

dtgiven that

dx

dt

∣∣∣∣y=125

= −12. From x2 + 102 = y2 we

get 2xdx

dt= 2y

dy

dtso

dy

dt=x

y

dx

dt. Use x2 + 100 = y2 to find that

x =√

15, 525 = 15√

69 when y = 125 sody

dt=

15√

69125

(−12) =

−36√

6925

. The rope must be pulled at the rate of36√

6925

ft/min.

y

x Boat

Pulley

10


64 Chapter 3

33. Finddx

dt

∣∣∣∣θ=π/4

given thatdθ

dt=

2π10

=π

5rad/s.

Then x = 4 tan θ (see figure) sodx

dt= 4 sec2 θ

dθ

dt,

dx

dt

∣∣∣∣θ=π/4

= 4(

sec2 π

4

)(π5

)= 8π/5 km/s.

x

4

Ship

!

35. We wish to finddz

dt

∣∣∣∣x=2,y=4

givendx

dt= −600 and

dy

dt

∣∣∣∣x=2,y=4

= −1200 (see figure). From the law of cosines,

z2 = x2 + y2 − 2xy cos 120◦ = x2 + y2 − 2xy(−1/2)

= x2+y2+xy, so 2zdz

dt= 2x

dx

dt+2y

dy

dt+x

dy

dt+y

dx

dt,

dz

dt=

12z

[(2x+ y)

dx

dt+ (2y + x)

dy

dt

].

AircraftP

Missile

x

yz

120º

When x = 2 and y = 4, z2 = 22 + 42 + (2)(4) = 28, so z =√

28 = 2√

7, thus

dz

dt

∣∣∣∣x=2,y=4

=1

2(2√

7)[(2(2) + 4)(−600) + (2(4) + 2)(−1200)] = −4200√

7= −600

√7 mi/h;

the distance between missile and aircraft is decreasing at the rate of 600√

7 mi/h.

37. (a) We wantdy

dt

∣∣∣∣x=1,y=2

given thatdx

dt

∣∣∣∣x=1,y=2

= 6. For convenience, first rewrite the equation as

xy3 =85

+85y2 then 3xy2 dy

dt+ y3 dx

dt=

165ydy

dt,dy

dt=

y3

165y − 3xy2

dx

dtso

dy

dt

∣∣∣∣x=1,y=2

=23

165

(2)− 3(1)22(6) = −60/7 units/s.

(b) falling, becausedy

dt< 0

39. The coordinates of P are (x, 2x), so the distance between P and the point (3, 0) is

D =√

(x− 3)2 + (2x− 0)2 =√

5x2 − 6x+ 9. FinddD

dt

∣∣∣∣x=3

given thatdx

dt

∣∣∣∣x=3

= −2.

dD

dt=

5x− 3√5x2 − 6x+ 9

dx

dt, so

dD

dt

∣∣∣∣x=3

=12√36

(−2) = −4 units/s.

41. Solvedx

dt= 3

dy

dtgiven y = x/(x2 + 1). Then y(x2 + 1) = x. Differentiating with respect to x,

(x2 + 1)dy

dx+ y(2x) = 1. But

dy

dx=dy/dt

dx/dt=

13

so (x2 + 1)13

+ 2xy = 1, x2 + 1 + 6xy = 3,

x2 + 1 + 6x2/(x2 + 1) = 3, (x2 + 1)2 + 6x2 − 3x2 − 3 = 0, x4 + 5x2 − 3 = 0. By the QuadraticFormula applied to x2 we obtain x2 = (−5 ±

√25 + 12)/2. The minus sign is spurious since x2

cannot be negative, so x2 = (√

37− 5)/2, x ≈ ±0.7357861545, y = ±− 0.4773550654.


Exercise Set 3.5 65

43. FinddS

dt

∣∣∣∣s=10

given thatds

dt

∣∣∣∣s=10

= −2. From1s

+1S

=16

we get − 1s2

ds

dt− 1S2

dS

dt= 0, so

dS

dt= −S

2

s2

ds

dt. If s = 10, then

110

+1S

=16

which gives S = 15. SodS

dt

∣∣∣∣s=10

= −225100

(−2) = 4.5

cm/s. The image is moving away from the lens.

45. Let r be the radius, V the volume, and A the surface area of a sphere. Show thatdr

dtis a constant

given thatdV

dt= −kA, where k is a positive constant. Because V =

43πr3,

dV

dt= 4πr2 dr

dt(1)

But it is given thatdV

dt= −kA or, because A = 4πr2,

dV

dt= −4πr2k which when substituted into

equation (1) gives −4πr2k = 4πr2 dr

dt,dr

dt= −k.

Extend sides of cup to complete the cone and let V0 be the volume

of the portion added, then (see figure) V =13πr2h− V0 where

r

h=

412

=13

so r =13h and V =

13π

(h

3

)2h − V0 =

127πh3 − V0,

dV

dt=

19πh2 dh

dt,dh

dt=

9πh2

dV

dt,dh

dt

∣∣∣∣h=9

=9

π(9)2(20) =

209π

cm/s.

r

4

2h

6

6

EXERCISE SET 3.5

1. (a) f(x) ≈ f(1) + f ′(1)(x− 1) = 1 + 3(x− 1)(b) f(1 + ∆x) ≈ f(1) + f ′(1)∆x = 1 + 3∆x(c) From Part (a), (1.02)3 ≈ 1 + 3(0.02) = 1.06. From Part (b), (1.02)3 ≈ 1 + 3(0.02) = 1.06.

3. (a) f(x) ≈ f(x0) + f ′(x0)(x − x0) = 1 + (1/(2√

1)(x − 0) = 1 + (1/2)x, so with x0 = 0 andx = −0.1, we have

√0.9 = f(−0.1) ≈ 1 + (1/2)(−0.1) = 1 − 0.05 = 0.95. With x = 0.1 we

have√

1.1 = f(0.1) ≈ 1 + (1/2)(0.1) = 1.05.(b) y

x

dy

0.1–0.1

!y !ydy

5. f(x) = (1+x)15 and x0 = 0. Thus (1+x)15 ≈ f(x0)+f ′(x0)(x−x0) = 1+15(1)14(x−0) = 1+15x.

7. tanx ≈ tan(0) + sec2(0)(x− 0) = x

9. x0 = 0, f(x) = ex, f ′(x) = ex, f ′(x0) = 1, hence ex ≈ 1 + 1 · x = 1 + x

11. x4 ≈ (1)4 + 4(1)3(x− 1). Set ∆x = x− 1; then x = ∆x+ 1 and (1 + ∆x)4 = 1 + 4∆x.


66 Chapter 3

13.1

2 + x≈ 1

2 + 1− 1

(2 + 1)2(x− 1), and 2 + x = 3 + ∆x, so

13 + ∆x

≈ 13− 1

9∆x

15. Let f(x) = tan−1 x, f(1) = π/4, f ′(1) = 1/2, tan−1(1 + ∆x) ≈ π

4+

12

∆x

17. f(x) =√x+ 3 and x0 = 0, so

√x+ 3 ≈

√3 +

12√

3(x− 0) =

√3 +

12√

3x, and∣∣∣∣f(x)−

(√3 +

12√

3x

)∣∣∣∣ < 0.1 if |x| < 1.692.

0

-0.1

-2 2

| f (x) – ( 3 + x)|12 3

19. tan 2x ≈ tan 0 + (sec2 0)(2x− 0) = 2x,and | tan 2x− 2x| < 0.1 if |x| < 0.3158

f x 2x

21. (a) The local linear approximation sinx ≈ x gives sin 1◦ = sin(π/180) ≈ π/180 = 0.0174533 anda calculator gives sin 1◦ = 0.0174524. The relative error | sin(π/180)−(π/180)|/(sinπ/180) =0.000051 is very small, so for such a small value of x the approximation is very good.

(b) Use x0 = 45◦ (this assumes you know, or can approximate,√

2/2).

(c) 44◦ =44π180

radians, and 45◦ =45π180

=π

4radians. With x =

44π180

and x0 =π

4we obtain

sin 44◦ = sin44π180

≈ sinπ

4+(

cosπ

4

)(44π180− π

4

)=√

22

+√

22

(−π180

)= 0.694765. With a

calculator, sin 44◦ = 0.694658.

23. f(x) = x4, f ′(x) = 4x3, x0 = 3, ∆x = 0.02; (3.02)4 ≈ 34 + (108)(0.02) = 81 + 2.16 = 83.16

25. f(x) =√x, f ′(x) =

12√x

, x0 = 64, ∆x = 1;√

65 ≈√

64 +116

(1) = 8 +116

= 8.0625

27. f(x) =√x, f ′(x) =

12√x

, x0 = 81, ∆x = −0.1;√

80.9 ≈√

81 +118

(−0.1) ≈ 8.9944

29. f(x) = sinx, f ′(x) = cosx, x0 = 0, ∆x = 0.1; sin 0.1 ≈ sin 0 + (cos 0)(0.1) = 0.1

31. f(x) = cosx, f ′(x) = − sinx, x0 = π/6, ∆x = π/180;

cos 31◦ ≈ cos 30◦ +(−1

2

)( π

180

)=√

32− π

360≈ 0.8573

33. tan−1(1 + ∆x) ≈ π

4+

12

∆x,∆x = −0.01, tan−1 0.99 ≈ π

4− 0.005 ≈ 0.780398

35. 3√

8.24 = 81/3 3√

1.03 ≈ 2(1 + 130.03) ≈ 2.02

4.083/2 = 43/21.023/2 = 8(1 + 0.02(3/2)) = 8.24


Exercise Set 3.5 67

37. (a) dy = f ′(x)dx = 2xdx = 4(1) = 4 and∆y = (x+ ∆x)2 − x2 = (2 + 1)2 − 22 = 5

(b)

x

y

dy !y

4

2 3

39. dy = 3x2dx;∆y = (x+ ∆x)3 − x3 = x3 + 3x2∆x+ 3x(∆x)2 + (∆x)3 − x3 = 3x2∆x+ 3x(∆x)2 + (∆x)3

41. dy = (2x− 2)dx;∆y= [(x+ ∆x)2 − 2(x+ ∆x) + 1]− [x2 − 2x+ 1]

= x2 + 2x ∆x+ (∆x)2 − 2x− 2∆x+ 1− x2 + 2x− 1 = 2x ∆x+ (∆x)2 − 2∆x

43. (a) dy = (12x2 − 14x)dx(b) dy = x d(cosx) + cosx dx = x(− sinx)dx+ cosxdx = (−x sinx+ cosx)dx

45. (a) dy =(√

1− x− x

2√

1− x

)dx =

2− 3x2√

1− xdx

(b) dy = −17(1 + x)−18dx

47. false; dy = (dy/dx)dx

49. false; they are equal whenever the function is linear

51. dy =3

2√

3x− 2dx, x = 2, dx = 0.03; ∆y ≈ dy =

34

(0.03) = 0.0225

53. dy =1− x2

(x2 + 1)2dx, x = 2, dx = −0.04; ∆y ≈ dy =

(− 3

25

)(−0.04) = 0.0048

55. (a) A = x2 where x is the length of a side; dA = 2x dx = 2(10)(±0.1) = ±2 ft2.

(b) relative error in x is ≈ dx

x=±0.110

= ±0.01 so percentage error in x is ≈ ±1%; relative error

in A is ≈ dA

A=

2x dxx2

= 2dx

x= 2(±0.01) = ±0.02 so percentage error in A is ≈ ±2%

57. (a) x = 10 sin θ, y = 10 cos θ (see figure),

dx= 10 cos θdθ = 10(

cosπ

6

)(± π

180

)= 10

(√3

2

)(± π

180

)≈ ±0.151 in,

dy= −10(sin θ)dθ = −10(

sinπ

6

)(± π

180

)= −10

(12

)(± π

180

)≈ ±0.087 in

10!x

y

"

(b) relative error in x is ≈ dx

x= (cot θ)dθ =

(cot

π

6

)(± π

180

)=√

3(± π

180

)≈ ±0.030

so percentage error in x is ≈ ±3.0%;

relative error in y is ≈ dy

y= − tan θdθ = −

(tan

π

6

)(± π

180

)= − 1√

3

(± π

180

)≈ ±0.010

so percentage error in y is ≈ ±1.0%


68 Chapter 3

59.dR

R=

(−2k/r3)dr(k/r2)

= −2dr

r, but

dr

r≈ ±0.05 so

dR

R≈ −2(±0.05) = ±0.10; percentage error in R

is ≈ ±10%

61. A =14

(4)2 sin 2θ = 4 sin 2θ thus dA = 8 cos 2θdθ so, with θ = 30◦ = π/6 radians and

dθ = ±15′ = ±1/4◦ = ±π/720 radians, dA = 8 cos(π/3)(±π/720) = ±π/180 ≈ ±0.017 cm2

63. V = x3 where x is the length of a side;dV

V=

3x2dx

x3= 3

dx

x, but

dx

x≈ ±0.02

sodV

V≈ 3(±0.02) = ±0.06; percentage error in V is ≈ ±6%.

65. A =14πD2 where D is the diameter of the circle;

dA

A=

(πD/2)dDπD2/4

= 2dD

D, but

dA

A≈ ±0.01 so

2dD

D≈ ±0.01,

dD

D≈ ±0.005; maximum permissible percentage error in D is ≈ ±0.5%.

67. V = volume of cylindrical rod = πr2h = πr2(15) = 15πr2; approximate ∆V by dV if r = 2.5 anddr = ∆r = 0.1. dV = 30πr dr = 30π(2.5)(0.1) ≈ 23.5619 cm3.

69. (a) α = ∆L/(L∆T ) = 0.006/(40× 10) = 1.5× 10−5/◦C(b) ∆L = 2.3× 10−5(180)(25) ≈ 0.1 cm, so the pole is about 180.1 cm long.

EXERCISE SET 3.6

1. (a) limx→2

x2 − 4x2 + 2x− 8

= limx→2

(x− 2)(x+ 2)(x+ 4)(x− 2)

= limx→2

x+ 2x+ 4

=23

, or, using L’Hopital’s Rule,

limx→2

x2 − 4x2 + 2x− 8

= limx→2

2x2x+ 2

=23

(b) limx→+∞

2x− 53x+ 7

=2− lim

x→+∞

5x

3 + limx→+∞

7x

=23

or, using L’Hopital’s Rule,

limx→+∞

2x− 53x+ 7

= limx→+∞

23

=23

3. true; lnx is not defined for negative x 5. false

7. limx→0

ex

cosx= 1 9. lim

θ→0

sec2 θ

1= 1

11. limx→π+

cosx1

= −1 13. limx→+∞

1/x1

= 0

15. limx→0+

− csc2 x

1/x= limx→0+

−xsin2 x

= limx→0+

−12 sinx cosx

= −∞

17. limx→+∞

100x99

ex= limx→+∞

(100)(99)x98

ex= · · · = lim

x→+∞

(100)(99)(98) · · · (1)ex

= 0

19. limx→0

2/√

1− 4x2

1= 2 21. lim

x→+∞xe−x = lim

x→+∞

x

ex= limx→+∞

1ex

= 0


Exercise Set 3.6 69

23. limx→+∞

x sin(π/x) = limx→+∞

sin(π/x)1/x

= limx→+∞

(−π/x2) cos(π/x)−1/x2

= limx→+∞

π cos(π/x) = π

25. limx→(π/2)−

sec 3x cos 5x = limx→(π/2)−

cos 5xcos 3x

= limx→(π/2)−

−5 sin 5x−3 sin 3x

=−5(+1)

(−3)(−1)= −5

3

27. y = (1− 3/x)x, limx→+∞

ln y = limx→+∞

ln(1− 3/x)1/x

= limx→+∞

−31− 3/x

= −3, limx→+∞

y = e−3

29. y = (ex + x)1/x, limx→0

ln y = limx→0

ln(ex + x)x

= limx→0

ex + 1ex + x

= 2, limx→0

y = e2

31. y = (2− x)tan(πx/2), limx→1

ln y = limx→1

ln(2− x)cot(πx/2)

= limx→1

2 sin2(πx/2)π(2− x)

= 2/π, limx→1

y = e2/π

33. limx→0

(1

sinx− 1x

)= limx→0

x− sinxx sinx

= limx→0

1− cosxx cosx+ sinx

= limx→0

sinx2 cosx− x sinx

= 0

35. limx→+∞

(x2 + x)− x2

√x2 + x+ x

= limx→+∞

x√x2 + x+ x

= limx→+∞

1√1 + 1/x+ 1

= 1/2

37. limx→+∞

[x− ln(x2 + 1)] = limx→+∞

[ln ex − ln(x2 + 1)] = limx→+∞

lnex

x2 + 1,

limx→+∞

ex

x2 + 1= limx→+∞

ex

2x= limx→+∞

ex

2= +∞ so lim

x→+∞[x− ln(x2 + 1)] = +∞

39. y = xsin x, ln y = sinx lnx, limx→0+

ln y = limx→0+

lnxcscx

= limx→0+

1/x− cscx cotx

= limx→0+

(sinxx

)(− tanx) =

1(−0) = 0, so limx→0+

xsin x = limx→0+

y = e0 = 1.

41. y =[− 1

lnx

]x, ln y = x ln

[− 1

ln x

],

limx→0+

ln y = limx→0+

ln[− 1

ln x

]1/x

= limx→0+

(− 1x lnx

)(−x2) = − lim

x→0+

x

lnx= 0, lim

x→0+y = e0 = 1

43. y = (lnx)1/x, ln y = (1/x) ln lnx,

limx→+∞

ln y = limx→+∞

ln lnxx

= limx→+∞

1/(x lnx)1

= 0, limx→+∞

y = 1

45. y = (tanx)π/2−x, ln y = (π/2− x) ln tanx, limx→(π/2)−

ln y = limx→(π/2)−

ln tanx1/(π/2− x)

= limx→(π/2)−

(sec2 x/ tanx)1/(π/2− x)2

= limx→(π/2)−

(π/2− x)cosx

(π/2− x)sinx

= limx→(π/2)−

(π/2− x)cosx

limx→(π/2)−

(π/2− x)sinx

= 1 · 0 = 0, limx→(π/2)−

y = 1

47. (a) L’Hopital’s Rule does not apply to the problem limx→1

3x2 − 2x+ 13x2 − 2x

because it is not a00

form.

(b) limx→1

3x2 − 2x+ 13x2 − 2x

= 2


70 Chapter 3

49. limx→+∞

1/(x lnx)1/(2√x)

= limx→+∞

2√x lnx

= 0 0.15

0100 10000

51. y = (sinx)3/ ln x,

limx→0+

ln y = limx→0+

3 ln sinxlnx

= limx→0+

(3 cosx)x

sinx= 3,

limx→0+

y = e3

25

190 0.5

53. lnx− ex = lnx− 1e−x

=e−x lnx− 1

e−x;

limx→+∞

e−x lnx = limx→+∞

lnxex

= limx→+∞

1/xex

= 0 by L’Hopital’s

Rule,

so limx→+∞

[lnx− ex] = limx→+∞

e−x lnx− 1e−x

= −∞

0

–16

0 3

55. y = (lnx)1/x,

limx→+∞

ln y = limx→+∞

ln(lnx)x

= limx→+∞

1x lnx

= 0;

limx→+∞

y = 1, y = 1 is the horizontal asymptote

1.02

1100 10000

57. (a) 0 (b) +∞ (c) 0 (d) −∞ (e) +∞ (f) −∞

59. limx→+∞

1 + 2 cos 2x1

does not exist, nor is it ±∞; limx→+∞

x+ sin 2xx

= limx→+∞

(1 +

sin 2xx

)= 1

61. limx→+∞

(2 + x cos 2x+ sin 2x) does not exist, nor is it ±∞; limx→+∞

x(2 + sin 2x)x+ 1

= limx→+∞

2 + sin 2x1 + 1/x

,

which does not exist because sin 2x oscillates between −1 and 1 as x→ +∞

63. limR→0+

V tL e−Rt/L

1=V t

L

65. (b) limx→+∞

x(k1/x − 1) = limt→0+

kt − 1t

= limt→0+

(ln k)kt

1= ln k

(c) ln 0.3 = −1.20397, 1024(

1024√

0.3− 1)

= −1.20327;

ln 2 = 0.69315, 1024(

1024√

2− 1)

= 0.69338


Review Exercises, Chapter 3 71

67. (a) No; sin(1/x) oscillates as x→ 0. (b) 0.05

–0.05

–0.35 0.35

(c) For the limit as x→ 0+ use the Squeezing Theorem together with the inequalities−x2 ≤ x2 sin(1/x) ≤ x2. For x→ 0− do the same; thus lim

x→0f(x) = 0.

69. limx→0+

sin(1/x)(sinx)/x

, limx→0+

sinxx

= 1 but limx→0+

sin(1/x) does not exist because sin(1/x) oscillates between

−1 and 1 as x→ +∞, so limx→0+

x sin(1/x)sinx

does not exist.

REVIEW EXERCISES, CHAPTER 3

1. (a) 3x2 + xdy

dx+ y − 2 = 0,

dy

dx=

2− y − 3x2

x(b) y = (1 + 2x− x3)/x = 1/x+ 2− x2, dy/dx = −1/x2 − 2x

(c)dy

dx=

2− (1/x+ 2− x2)− 3x2

x= −1/x2 − 2x

3. − 1y2

dy

dx− 1x2

= 0 sody

dx= −y

2

x2

5.(xdy

dx+ y

)sec(xy) tan(xy) =

dy

dx,dy

dx=

y sec(xy) tan(xy)1− x sec(xy) tan(xy)

7.dy

dx=

3x4y

,d2y

dx2=

(4y)(3)− (3x)(4dy/dx)16y2

=12y − 12x(3x/(4y))

16y2=

12y2 − 9x2

16y3=−3(3x2 − 4y2)

16y3,

but 3x2 − 4y2 = 7 sod2y

dx2=−3(7)16y3

= − 2116y3

9.dy

dx= tan(πy/2) + x(π/2)

dy

dxsec2(πy/2),

dy

dx

]y=1/2

= 1 + (π/4)dy

dx

]y=1/2

(2),dy

dx

]y=1/2

=2

2− π

11. Substitute y = mx into x2 +xy+ y2 = 4 to get x2 +mx2 +m2x2 = 4, which has distinct solutionsx = ±2/

√m2 +m+ 1. They are distinct because m2 + m + 1 = (m + 1/2)2 + 3/4 ≥ 3/4, so

m2 +m+ 1 is never zero.

Note that the points of intersection occur in pairs (x0, y0) and (−x0,−y0). By implicit differenti-ation, the slope of the tangent line to the ellipse is given bydy/dx = −(2x + y)/(x + 2y). Since the slope is unchanged if we replace (x, y) with (−x,−y), itfollows that the slopes are equal at the two point of intersection.Finally we must examine the special case x = 0 which cannot be written in the form y = mx. Ifx = 0 then y = ±2, and the formula for dy/dx gives dy/dx = −1/2, so the slopes are equal.


72 Chapter 3

13. By implicit differentiation, 3x2 − y− xy′ + 3y2y′ = 0, so y′ = (3x2 − y)/(x− 3y2). This derivativeexists except when x = 3y2. Substituting this into the original equation x3− xy+ y3 = 0, one has27y6 − 3y3 + y3 = 0, y3(27y3 − 2) = 0. The unique solution in the first quadrant isy = 21/3/3, x = 3y2 = 22/3/3

15. y = ln(x+ 1) + 2 ln(x+ 2)− 3 ln(x+ 3)− 4 ln(x+ 4), dy/dx =1

x+ 1+

2x+ 2

− 3x+ 3

− 4x+ 4

17.1

2x(2) = 1/x 19.

13x(lnx+ 1)2/3

21. log10 lnx =ln lnxln 10

, y′ =1

(ln 10)(x lnx)

23. y =32

lnx+12

ln(1 + x4), y′ =3

2x+

2x3

(1 + x4)25. y = x2 + 1 so y′ = 2x.

27. y′ = 2e√x + 2xe

√x d

dx

√x = 2e

√x +√xe√x 29. y′ =

2π(1 + 4x2)

31. ln y = ex lnx,y′

y= ex

(1x

+ lnx)

,dy

dx= xe

x

ex(

1x

+ lnx)

= ex[xe

x−1 + xex

lnx]

33. y′ =2

|2x+ 1|√

(2x+ 1)2 − 1

35. ln y = 3 lnx− 12

ln(x2 + 1), y′/y =3x− x

x2 + 1, y′ =

3x2

√x2 + 1

− x4

(x2 + 1)3/2

37. (b) y

x

2

4

6

1 2 3 4

(c)dy

dx=

12− 1x

sody

dx< 0 at x = 1 and

dy

dx> 0 at x = e

(d) The slope is a continuous function which goes from a negative value to a positive value;therefore it must take the value zero between, by the Intermediate Value Theorem.

(e)dy

dx= 0 when x = 2

39. Solvedy

dt= 3

dx

dtgiven y = x lnx. Then

dy

dt=dy

dx

dx

dt= (1 + lnx)

dx

dt, so 1 + lnx = 3, lnx = 2,

x = e2.

41. Set y = logb x and solve y′ = 1: y′ =1

x ln b= 1 so x =

1ln b

. The curves

intersect when (x, x) lies on the graph of y = logb x, so x = logb x.

From Formula (8), Section 1.6, logb x =lnxln b

from which lnx = 1,

x = e, ln b = 1/e, b = e1/e ≈ 1.4447.

y

x

2

2


Review Exercises, Chapter 3 73

43. Yes, g must be differentiable (where f ′ 6= 0); this can be inferred from the graphs. Note that iff ′ = 0 at a point then g′ cannot exist (infinite slope).

45. Let P (x0, y0) be a point on y = e3x then y0 = e3x0 . dy/dx = 3e3x so mtan = 3e3x0 at P and anequation of the tangent line at P is y − y0 = 3e3x0(x − x0), y − e3x0 = 3e3x0(x − x0). If the linepasses through the origin then (0, 0) must satisfy the equation so −e3x0 = −3x0e

3x0 which givesx0 = 1/3 and thus y0 = e. The point is (1/3, e).

47. ln y = 2x ln 3 + 7x ln 5;dy/dx

y= 2 ln 3 + 7 ln 5, or

dy

dx= (2 ln 3 + 7 ln 5)y

49. y′= aeax sin bx+ beax cos bx and y′′ = (a2 − b2)eax sin bx+ 2abeax cos bx, so y′′ − 2ay′ + (a2 + b2)y= (a2 − b2)eax sin bx+ 2abeax cos bx− 2a(aeax sin bx+ beax cos bx) + (a2 + b2)eax sin bx = 0.

51. (a) 100

200 8

(b) as t tends to +∞, the population tends to 19

limt→+∞

P (t) = limt→+∞

955− 4e−t/4

=95

5− 4 limt→+∞

e−t/4=

955

= 19

(c) the rate of population growth tends to zero 0

–80

0 8

53. In the case +∞ − (−∞) the limit is +∞; in the case −∞ − (+∞) the limit is −∞, becauselarge positive (negative) quantities are added to large positive (negative) quantities. The cases+∞− (+∞) and −∞− (−∞) are indeterminate; large numbers of opposite sign are subtracted,and more information about the sizes is needed.

55. limx→+∞

(ex − x2) = limx→+∞

x2(ex/x2 − 1), but limx→+∞

ex

x2= limx→+∞

ex

2x= limx→+∞

ex

2= +∞

so limx→+∞

(ex/x2 − 1) = +∞ and thus limx→+∞

x2(ex/x2 − 1) = +∞

57. limx→0

x2ex

sin2 3x=[

limx→0

3xsin 3x

]2ex

9=

19

59. A = πr2 anddr

dt= −5, so

dA

dt=dA

dr

dr

dt= 2πr(−5) = −500π, so the area is shrinking at a rate of

500πm2/min.


74 Chapter 3

61. (a) ∆x = 1.5− 2 = −0.5; dy =−1

(x− 1)2∆x =

−1(2− 1)2

(−0.5) = 0.5; and

∆y =1

(1.5− 1)− 1

(2− 1)= 2− 1 = 1.

(b) ∆x = 0− (−π/4) = π/4; dy =(sec2(−π/4)

)(π/4) = π/2; and ∆y = tan 0− tan(−π/4) = 1.

(c) ∆x = 3− 0 = 3; dy =−x√

25− x2=

−0√25− (0)2

(3) = 0; and

∆y =√

25− 32 −√

25− 02 = 4− 5 = −1.

63. (a) h = 115 tanφ, dh = 115 sec2 φdφ; with φ = 51◦ =51180

π radians and

dφ = ±0.5◦ = ±0.5( π

180

)radians, h ± dh = 115(1.2349) ± 2.5340 = 142.0135 ± 2.5340, so

the height lies between 139.48 m and 144.55 m.

(b) If |dh| ≤ 5 then |dφ| ≤ 5115

cos2 51180

π ≈ 0.017 radian, or ldφ| ≤ 0.98◦.

MAKING CONNECTIONS, CHAPTER 3

1. (a) If t > 0 then A(−t) is the amount K there was t time-units ago in order that there be 1 unit

now, i.e. K ·A(t) = 1, so K =1

A(t). But, as said above, K = A(−t). So A(−t) =

1A(t)

.

(b) (i) If s and t are positive, then the amount 1 becomes A(s) after s seconds, and that in turnis A(s)A(t) after another t seconds, i.e. 1 becomes A(s)A(t) after s + t seconds. But thisamount is also A(s+ t), so A(s)A(t) = A(s+ t).(ii) If 0 ≤ −s ≤ t then A(−s)A(s+ t) = A(t). From Part (i) we get A(s+ t) = A(s)A(t).

(iii) If 0 ≤ t ≤ −s then A(s+ t) =1

A(−s− t)=

1A(−s)A(−t)

= A(s)A(t)

by Parts (i) and (ii).(iv) If s and t are both negative then by Part (i),

A(s+ t) =1

A(−s− t)=

1A(−s)A(−t)

= A(s)A(t).

(c) If n > 0 then A

(1n

)A

(1n

). . . A

(1n

)= A

(n

1n

)= A(1), so A

(1n

)= A(1)1/n = b1/n

from Part (b). If n < 0 then by Part (a), A(

1n

)=

1A(− 1n

) =1

A(1)−1/n= A(1)1/n = b1/n

(d) Let m,n be integers. Assume n 6= 0 and m > 0. Then A(mn

)= A

(1n

)m= A(1)m/n = bm/n

(e) If f, g are continuous functions of t and f and g are equal on the rational numbers{mn

: n 6= 0}

,

then f(t) = g(t) for all t. Because if x is irrational, then let tn be a sequence of rational num-bers which converges to x. Then for all n > 0, f(tn) = g(tn) and thus f(x) = lim

n→+∞f(tn) =

limn→+∞

g(tn) = g(x)

mac2312 ch 3 solutions

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